Classifying $\mathbb{Z} \times \mathbb{Z} \times \mathbb{Z} / \langle(3,3,3)\rangle$ Again, my instructor's explanation in class is so confusing to me.  Here we cannot take the original approach of finding the order and limiting our choices.  So instead my instructor says that this is isomorphic to $\mathbb{Z} \times \mathbb{Z} \times \mathbb{Z}_3$.  To justify it he said that $\mathbb{Z} \times \mathbb{Z} \times \mathbb{Z}$ is 3 dimensional and the denominator is one dimensional but this doesn't quite explain where the $\mathbb{Z}_3$ comes from?  Can anyone provide a better way to think about this?  It is making no sense to me at all.  
 A: We have the abelian group with the presentation $\langle a,b,c : 3a+3b+3c=0\rangle$. Since $3(a+b+c)=0$ in this group, there is a homomorphism 
$$\langle a,b,x : 3x=0 \rangle \to \langle a,b,c : 3a+3b+3c=0\rangle$$
mapping $a \mapsto a$, $b \mapsto b$, $x \mapsto a+b+c$. It is an isomorphism since it has an inverse homomorphism defined by $a \mapsto a, b \mapsto b, c \mapsto x-a-b$. But $\langle a,b,x : 3x=0 \rangle \cong \mathbb{Z} \times \mathbb{Z} \times \mathbb{Z}/3$.
A: You're asking for the quotient of $\mathbf{Z}^3$ by the rowspace of the matrix
$$ \left( \begin{matrix} 3 & 3 & 3 \end{matrix} \right) $$
There's nothing we can do with integer row operations to simplify it that way, but column operations correspond to a change-of-basis transformation on $\mathbf{Z}^3$, and we can easily simplify this matrix with column operations to
$$ \left( \begin{matrix} 3 & 0 & 0 \end{matrix} \right) $$
and it's easy to see what the quotient of $\mathbf{Z}^3$ by $(3,0,0)$ is!
A: Think about it the following way: Any element in $Z \times Z \times Z$ has the form $(a,b,c)$. Write $c=3q+r$ with $0 \leq r \leq 2$. 
Now since $(3q,3q,3q) \in <(3,3,3)>$ it follows that in $Z \times Z \times Z /\langle (3,3,3)\rangle$ we have
$$(a,b,c) \equiv (a-3q, b-3q, r)$$
Now, if you use long division you will see that this identification is compatible with addition in $Z \times Z \times Z_3$, and it is easy to define now an isomorphism.
A: Consider the map \begin{equation}
\mathbb{Z} \times \mathbb{Z} \times \mathbb {Z} \longrightarrow\mathbb{Z} \times \mathbb{Z} \times \mathbb{Z}/{3\mathbb{Z}} 
\end{equation} 
$
  (x,y,z) \mapsto (x-y, z-y, y \hspace{1mm}\text{mod} \hspace{1 mm} 3)$. You should be able to see kernel is generated by $(3,3,3)$. 
A: The set $\{v_1, v_2, v_3\} = \{(1,0,0), (0,1,0), (1,1,1)\}$ is a $\mathbb{Z}$-basis for $\mathbb{Z} \times \mathbb{Z} \times \mathbb{Z}$, so $\mathbb{Z} \times \mathbb{Z} \times \mathbb{Z} = \mathbb{Z} v_1 \oplus \mathbb{Z} v_2 \oplus \mathbb{Z} v_3$.  Then
$$
\frac{\mathbb{Z} \times \mathbb{Z} \times \mathbb{Z}}{\langle (3,3,3)\rangle} = \frac{\mathbb{Z} v_1 \oplus \mathbb{Z} v_2 \oplus \mathbb{Z} v_3}{\mathbb{Z} 3 v_3} \cong \mathbb{Z} v_1 \oplus \mathbb{Z} v_2 \oplus \frac{\mathbb{Z} v_3}{\mathbb{Z} 3 v_3} \cong \mathbb{Z} \oplus \mathbb{Z} \oplus \frac{\mathbb{Z}}{3 \mathbb{Z}} \, .
$$
